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Kitty [74]
3 years ago
9

For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.

Engineering
1 answer:
horrorfan [7]3 years ago
5 0

Answer:

attached below

Explanation:

a) G(s) = 1 / s( s+2)(s + 4 )

Bode asymptotic magnitude and asymptotic phase plots

attached below

b) G(s) = (s+5)/(s+2)(s+4)

phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4

attached below

c) G(s)= (s+3)(s+5)/s(s+2)(s+4)

solution attached below

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Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep t
HACTEHA [7]

Answer:

b)1.08 N

Explanation:

Given that

velocity of air V= 45 m/s

Diameter of pipe = 2 cm

Force exerted by fluid  F

F=\rho AV^2

So force exerted in x-direction

F_x=\rho AV^2

F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

F=0.763 N

So force exerted in y-direction

F_y=\rho AV^2

F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

F=0.763 N

So the resultant force R

R=\sqrt{F_x^2+F_y^2}

R=\sqrt{0.763^2+0.763^2}

R=1.079

So the force required to hold the pipe is 1.08 N.

3 0
3 years ago
Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale
artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

  • HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ \longrightarrow 2Cl₂ + 2H₂O

Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \  of \ limiting \ reactant \ supplied \ in \ the \, feed}}

Which gives;

80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

  • <u>HCl: 0.4 kmol/h</u>
  • <u>Cl₂: 1.6 kmol/h</u>
  • <u>H₂O: 1.6 kmol/h</u>
  • <u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

brainly.com/question/14668990

brainly.com/question/82989

7 0
3 years ago
Brainly and points if you want
Tju [1.3M]

Answer:

thank you

Explanation:

have a nice day

8 0
3 years ago
Read 2 more answers
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5
Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

3 0
3 years ago
What do you own that might not be manufactured?
horrorfan [7]

Answer:

A pet

Explanation:

Latin time I checked animals aren't made by people? I honestly don't know if this helps but I'm technically not wrong.

8 0
3 years ago
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